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Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 23

Answers (1)

Answer:

 \frac{3}{5}

Hint:

 Use Baye’s theorem.

Given:

 From A, B, C chance of being selected as manager of firm in ratio 4:1:2 respectively. The respectively probability for them a radical change in marketing strategy, 0.3, 0.8, 0.5.

Solution:

Let E1 ,E2,E3  denote that event that the change take place is selected B is selected C is selected.

\begin{aligned} &P(E_1)=\frac{4}{7}\\ &P(E_2)=\frac{1}{37}\\ &P(E_3)=\frac{2}{7}\\ &P\left ( \frac{A}{E_1} \right )=0.3\\ &P\left ( \frac{A}{E_2} \right )=0.8\\ &P\left ( \frac{A}{E_3} \right )=0.5\\ \end{aligned}

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{4}{7}\times 0.3 }}{\frac{4}{7}\times 0.3+\frac{1}{7}\times 0.8+\frac{2}{7}\times 0.5}\\ &=\frac{1.2}{1.2+0.8+1}\\ &=\frac{1.2}{3}\\ &=\frac{12}{30}\\ &=\frac{2}{5}\\ &\text { Required probability }=1-P\left (\frac{E_1}{A} \right )=1-\frac{2}{5}=\frac{3}{5} \end{aligned}

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Gurleen Kaur

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